Re: uN = vN iff uv^-1 is in N??

If $uN=vN$ then $\exists\{s,t\}\subset N $ such that $us=vt$. So
$us=vt\\st^{-1}=u^{-1}v\\st^{-1}\in N$.
That is one way. You post the other,

Sorry I don't think I stated the problem properly.

Let G be a group and N be a normal subgroup of G, ie $\displaystyle N \unlhd G$. Then there is a theorem that states:
For $\displaystyle u, v \in G $ we have $\displaystyle uN = vN$ if and only if $\displaystyle v^{-1} u \in N $.

Now my question is that if the group G is non-abelian, is it also true that: $\displaystyle uN = vN$ if and only if $\displaystyle v u^{-1} \in N $

So what you have there is the proof of the original theorem, which I understand (both directions). But the problem is that I want to show that $\displaystyle v u^{-1} \in N $ which I can't seem to do because the inverse always seems to be on the left hand side...

Re: uN = vN iff uv^-1 is in N??

Hi,
The operative word is "normal subgroup". If N is a normal subgroup of G and $u,v\in G$,
$uN=vN$ if and only if $uv^{-1}\in N$. The proof is almost immediate from the fact that N is a normal subgroup if and only if $uN=Nu$ for all $u\in G$. Now if you have trouble with the proof, post your problems.

If N is not a normal subgroup, your assertion is false. See if you can't find an example in $S_3$, where this is the full symmetric group on 3 letters. Again, if you have trouble, post your difficulties.

Re: uN = vN iff uv^-1 is in N??

Originally Posted by johng

Hi,
The operative word is "normal subgroup". If N is a normal subgroup of G and $u,v\in G$,
$uN=vN$ if and only if $uv^{-1}\in N$. The proof is almost immediate from the fact that N is a normal subgroup if and only if $uN=Nu$ for all $u\in G$. Now if you have trouble with the proof, post your problems.

If N is not a normal subgroup, your assertion is false. See if you can't find an example in $S_3$, where this is the full symmetric group on 3 letters. Again, if you have trouble, post your difficulties.

As you can see, N(2 3)N(1 3) has 4 distinct elements, so it cannot possibly be ANY coset of N (all of which have just 2 elements).

So for that subgroup, there is no feasible way to define "coset multiplication".

***********

There is another way to look at this situation:

Suppose we define an equivalence relation ~ on a group G that respects the group multiplication: that is, if [g] is the equivalence class of g, whenever [g] = [g'] and [h] = [h'], we have [gh] = [g'h'].

We could then turn the set of equivalence classes into a group by setting: [g]*[h] = [gh].

I claim [e] is a subgroup of G. To see this, suppose a,b are in [e]. Then since [a] = [e] = [b], we have [ab-1] = [a]*[b-1] = [b]*[b-1] = [bb-1] = [e], so ab-1 is in [e].

Note this also show that if [a] = [b], that is if a ~ b, we have ab-1 in [e].

Finally, I claim that [e] is a NORMAL subgroup of G. For consider [ghg-1], for h in [e]. We have [ghg-1] = [g]*[h]*[g-1] = [g]*[e]*[g-1] = [geg-1] = [gg-1] = [e] = [h],

whence ghg-1h-1 is in [e], so that ghg-1 is in h[e], which is a subset of [h]*[e] = [e]*[e] = [ee] = [e].

***************

There is also a third way of viewing this state of affairs:

Suppose we have a homomorphism f:G-->G'. We can view this as a surjective (onto) homomorphism f:G-->f(G).

Let K = {g in G: f(g) = e'}, where e' is the identity of G'. We can define an equivalence relation ~ on G by:

g ~ g' if and only if f(g) = f(g'), and we then have: K = [e]. We now show this equivalence relation respects the multiplication of G:

Suppose f(g) = f(g'), and f(h) = f(h'). Since f is a homomorphism, f(gh) = f(g)f(h) = f(g')f(h') = f(g'h'). As we saw above, this means K is a subgroup of G (it may be instructive to try to show this directly).

Next, we show that a ~ b if and only if Ka = Kb, that is, ab-1 is in K.